integral from 1 to 4 of (-3)/((2x+1)^2)

Lösung

\int_{1}^{4} \frac{- 3}{( 2 x + 1 ) ^{2}} d x

- \frac{1}{3}

Dezimale

- 0.33333 \dots

Schritte zur Lösung

\int_{1}^{4} \frac{- 3}{( 2 x + 1 ) ^{2}} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 3 \cdot \int_{1}^{4} \frac{1}{( 2 x + 1 ) ^{2}} d x

Wende U-Substitution an

= - 3 \cdot \int_{3}^{9} \frac{1}{2 u ^{2}} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 3 \cdot \frac{1}{2} \cdot \int_{3}^{9} \frac{1}{u ^{2}} d u

Wende die Potenzregel an

= - 3 \cdot \frac{1}{2} [- \frac{1}{u}]_{3}^{9}

Vereinfache

- 3 \cdot \frac{1}{2} [- \frac{1}{u}]_{3}^{9} : - \frac{3}{2} [- \frac{1}{u}]_{3}^{9}

= - \frac{3}{2} [- \frac{1}{u}]_{3}^{9}

Berechne die Grenzen:

\frac{2}{9}

= - \frac{3}{2} \cdot \frac{2}{9}

Vereinfache

= - \frac{1}{3}

\int_{0}^{x} t^{6} 1 + t^{7} d t

\int_{4}^{1} 4 x + 3 x^{2} - 2 x^{3} d x

\int_{- a}^{a} (a^{2} - t^{2}) d t

\int_{0}^{4} (2 t + 1) d t

\int_{0}^{20} x + sin (x) d x